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The ratio of the densities of oxygen and nitrogen is 16:14. At what temperature, the velocity of sound in oxygen will be equal to its velocity in nitrogen at 14°C? |
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Answer» `45^(@)` C `v=sqrt((lambda RT)/M)` where `lambda` is the ratio of specific HEATS `(C_(p)/C_(V))` R is the universal gas constant T is the absolutee temperature and M is the molecular weight of the gas. Let the velocity of sound in oxygen at t°C be equal to the velocity of sound in nitrogen at `14^(@)` C Now, velocity of sound in oxygen at `t^(@) C, v_(0) = sqrt((lambdaR(t+273))/M_(0))` and velocity of sound in nitrogen at `14^(@)` C, `v_(N) = sqrt((gammaR(14 + 273))/M_(N))` The value of `gamma` for both the gases will be same as both are, diatomic. Now, according to the question, `v_(0) =v_(N)` or `sqrt((gamma R(14 + 273))/M_(N))` or `(t+273)/287 = M_(O)/M_(N)` But, ratio of the densities of oxygen and nitrogen is, `M_(O)/M_(N) = 16/14` `therefore (t + 273)/287 = 16/14 = 8/7` or `7t + 273 xx 7 = 8 xx 287` or `t = 55^(@)` C Therefore, the velocity of sound in oxygen at 55°C will be equal to the velocity of sound in nitrogen at 14°C. |
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